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3 years ago

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Twenty percent of all trucks undergoing a certain inspection will fail the inspection. Assume that 40 trucks are independently u

ndergoing this inspection, one at a time. The expected number of trucks who fail the inspection is: A.8 B.16 C.20 D.40

1 answer:

BaLLatris [955]3 years ago

4 0

i was going to answer but siomeome beat me lol but hdjwStep-by-step explanation:

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Write two verbal expressions for n-5

sukhopar [10]

A number less than five

OR

n minus five

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3 years ago

Find the area and perimeter.<br> The area is ... and the perimeter is ...

Schach [20]

Answer:

Area: 6a^2

Perimeter: 12a^2

Step-by-step explanation:

To find the area of a triangle, use the formula A=1/2 b*h. The base is 3a^2 and the height is 4a^2.

A= 1/2 (3a^2)(4a^2)

A = 1/2 (12a^4)

A = 6a^4

To find the perimeter of the triangle, add each of the sides of the triangle. The third side, the hypotenuse, can be found using Pythagorean theorem.

$a^2+b^2=c^2\\(3a^2)^2+(4a^2)^2 = c^2\\(9a^4)+(16a^4)=c^2\\25a^4=c^2\\5a^2=c$

Now add each of the sides: 3a^2 + 4a^2 +5a^2 = 12a^2

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3 years ago

HURRY ASAP PLZZZZ ANYONE???? There are 25 towns in a certain country, and every pair of them is connected by a train route. How

Julli [10]

Answer:

25 train routes

Step-by-step explanation:

The train routes go from 1-2, 2-3, 3-4, 4-5, ,and so on, but there is also one that goes from 25-1. If you notice, there when we get to 4-5, there are 4 train routes which means that if there are 25 towns, then there are 24 train routes, but since 25 and 1 are connected then there are 25 routes. Hope this helps!

4 0

3 years ago

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Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

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3 years ago

What is the volume of a cube whose edges each have a measure of 1.5 cm?

inysia [295]

Using the cube formula for volume I got V=3.38cm^3

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3 years ago

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